Bootstrapper said:
I guess a thermistor may work to keep dc bias in tune.
You need to bypass the choke with a diode to the cap bank. Once boosted, the diode is backbiased. And come to think of it with this method you do need a thermistor or some way to limit current so as not to blow the diode.
No.
The problem is when output voltage is 0 and input is applied. The current must not flow uncontrolled through the choke when the switching transistor is activated.
The problem also occurs on overload: Output voltage drops if the current limiting is working. When output voltage tries to go below input voltage the current will flow uncontrolledly through the choke possibly saturating it. If the transistor tries to switch in this state it might blow up.
The problem is when output voltage is 0 and input is applied. The current must not flow uncontrolled through the choke when the switching transistor is activated.
The problem also occurs on overload: Output voltage drops if the current limiting is working. When output voltage tries to go below input voltage the current will flow uncontrolledly through the choke possibly saturating it. If the transistor tries to switch in this state it might blow up.
megajocke said:
That's why I intend to use the thermistor at the input.
megajocke said:
This is naturally true, if the output diodes get forward biased. But I don't see how the diode would help with this kind of situation, will you try to explain more? This is a valuable point, I will consider some sort of overload/short circuit protection. Thanks!
Well, now I have this overvoltage protection circuit planned and simulated. It comprises a single, fast & rail-to-rail comparator which compares the output voltage to a fixed reference. Whenever output voltage rises above the threshold, the comparator turns on, and drives the controller compensation pin (UC2843) to low value via a bjt. That turns the FET frive off which prevents the output voltage from rising any further. Seems to work fine, but care must be taken when designing the layout.
Now I lack the overcurrent protection. It ain't much of a problem to sense it, but then what? Wait until the input fuse blows? Otherwise I should implement some sort of a high-current relay to disconnect the input source from the rest of the converter. I can design a circuit that senses the input current and when it raises above, say, 30 amps, the controller will be shut down to prevent fatal damage to switches. By then, the fuse should blow.
Any other view will be welcomed!
Now I lack the overcurrent protection. It ain't much of a problem to sense it, but then what? Wait until the input fuse blows? Otherwise I should implement some sort of a high-current relay to disconnect the input source from the rest of the converter. I can design a circuit that senses the input current and when it raises above, say, 30 amps, the controller will be shut down to prevent fatal damage to switches. By then, the fuse should blow.
Any other view will be welcomed!
Update: I read a bit more about the compensation and the result is shown in the attachment. Now I get an even better transient response - although it isn't optimized yet.
Still, I feel I can make this baby work pretty soon, and I will leave space for extra compensation/snubbing/filtering components on the PCB
I will publish all the necessary information here for others if I get a decent result.
Output voltage @ 10 A step load
Still, I feel I can make this baby work pretty soon, and I will leave space for extra compensation/snubbing/filtering components on the PCB
I will publish all the necessary information here for others if I get a decent result.Output voltage @ 10 A step load
I will change the values of low pass filter across the sense resistor (R12 and C7) in order to reduce power losses (I^2*R12).
The frequency (cut-off) with the actual values is around 1.59Mhz.
The input impedance of this network (assuming an armonic signal at its input) is Zlp=R12+XC7=20Ohm (where XC7=1/(2*pi()*f*C7)).
It is true that sense resistor (R6) is much lower than Zlp, but I will feel more comfortable with let's say R12=1kOhm.
Keeping the same cut-off frequency, the C7 value is found to be 100pF (C7=1/(2*pi()*R12*f)).
On the voltage error amplifier network - R13 can be skipped. The EA gain will be set only by (R11||C9) and R7+Rtrim. R8 does not enter into calculations since the EA is trying to keep the inverting input potential at the non-inverting input one.
Input stage - It's well known that the input current of a boost converter is not switched, still, are you thinking to consider an input filter?
Regards,
Florin
The frequency (cut-off) with the actual values is around 1.59Mhz.
The input impedance of this network (assuming an armonic signal at its input) is Zlp=R12+XC7=20Ohm (where XC7=1/(2*pi()*f*C7)).
It is true that sense resistor (R6) is much lower than Zlp, but I will feel more comfortable with let's say R12=1kOhm.
Keeping the same cut-off frequency, the C7 value is found to be 100pF (C7=1/(2*pi()*R12*f)).
On the voltage error amplifier network - R13 can be skipped. The EA gain will be set only by (R11||C9) and R7+Rtrim. R8 does not enter into calculations since the EA is trying to keep the inverting input potential at the non-inverting input one.
Input stage - It's well known that the input current of a boost converter is not switched, still, are you thinking to consider an input filter?
Regards,
Florin
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